If `I=int(e^x)/(e^(4x)+e^(2x)+1) dx. J=int(e^(-x))/(e^(-4x)+e^(-2x)+1) dx.` Then for an arbitrary constant c, the value of `J-I` equal to

Options

A. `1/2 log |(e^(4x)-e^(2x)+1)/(e^(4x)+e^(2x)+1)|+C`
B. `1/2 log |(e^(2x)+e^(x)+1)/(e^(2x)-e^(x)+1)|+C`
C. `1/2 log |(e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1)|+C`
D. `1/2 log |(e^(4x)+e^(2x)+1)/(e^(4x)-e^(2x)+1)|+C`

Answer

To solve the problem, we need to evaluate the integrals \( I \) and \( J \) and then find \( J - I \).

### Step 1: Define the integrals
Let
\[ I = \int \frac{e^x}{e^{4x} + e^{2x} + 1} \, dx \]
and
\[ J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} \, dx. \]

### Step 2: Simplify \( J \)
We can rewrite \( J \) as follows:
\[ J = \int \frac{1/e^x}{1/e^{4x} + 1/e^{2x} + 1} \, dx = \int \frac{1}{\frac{1}{e^x}(e^{4x} + e^{2x} + 1)} \, dx = \int \frac{e^x}{e^{4x} + e^{2x} + 1} \, dx. \]
Thus, we have:
\[ J = \int \frac{e^x}{e^{4x} + e^{2x} + 1} \, dx = I. \]

### Step 3: Find \( J - I \)
Since both integrals are equal, we have:
\[ J - I = I - I = 0. \]

### Conclusion
The value of \( J - I \) is:
\[ \boxed{0}. \]